Question

2^7x=83 using logarithmsss

Answer 1

\(\Large \color{MidnightBlue}{\Rightarrow log_{2}83 = 7x }\)
Do you know about logarithms?

Answer 2

\(\Large \color{MidnightBlue}{\Rightarrow x^n = y \Longleftrightarrow log_{x}y = n }\)

Answer 3

\[\LARGE x = \frac{\log_2 83}{7} = \frac{1}{7} \log_2 83 = \log_2 83^7\] just to make it easier

Answer 4

@lgbasallote \[\huge{\log_283^{\frac{1}{7}}}\]

Answer 5

oh yes...thanks for the correction

Answer 6

\[\LARGE \log_2 83^{\frac{1}{7}} = \log_2 \sqrt[7]{83}\]

Answer 7

I have to solve for x...im not sure if this is right but i have [x=\log83-7\log_2/\log_2\] what do i do nextt??

Answer 8

\[x=\log83-7\log_2/\log_2\]

Answer 9

how do you read that?

Answer 10

yes, just like that

Answer 11

i don't get the \(7\log_2/\log_2\) part

Answer 12

log_2 is on the bottom the rest is the numerator

Answer 13

well it's impossible to just have \(7\log_2\) you need a power

Answer 14

let me just tell you that \[\LARGE \frac{\log_2 83}{7} \ne \log_2 (\frac{83}{7})\]

Answer 15

\[\LARGE \frac{\log_2 83}{7} = \log_2 83^{\frac{1}{7}}\] \[\mathbf{BUT}\] \[\LARGE \log_2 (\frac{83}{7}) = \log_2 83 - \log_2 7\]

therefore they are NOT equal

Answer 16

nooo lgbasallote, thats not the equation but i dont quite know how to write it on this
srinidhijha! That looks right but why did you write 7xLog_2, its not just 7xLog ?

Answer 17

i dont know how you got \(7\log_2\) :/

Answer 18

@Lovely95 bcz i have taken log on boyh sides with base 10 . If u dont write a base ex-2,e then it means base of log is 10

so 7xlog2 = (7x ) {log base 10 (2)} whichz not equal to (7x) log base 10

Answer 19

@srinidhijha But .228 doesnt check out :/

Answer 20

mistaken! this is the correct answer you can do it by taking base 2 . its simpler and answer is CORRECT for sure

Answer 21

Yes! its checks THANK YOUU